2012 AMC 12A Problem 18

Below is the professionally curated solution for Problem 18 of the 2012 AMC 12A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2012 AMC 12A solutions, or check the answer key.

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Concepts:incircle, incenter, and inradiusHeron’s FormulaPythagorean Theorem

Difficulty rating: 1980

18.

Triangle ABCABC has AB=27,AB = 27, AC=26,AC = 26, and BC=25.BC = 25. Let II denote the intersection of the internal angle bisectors of ABC.\triangle ABC. What is BI?BI?

1515

5+26+335 + \sqrt{26} + 3\sqrt{3}

3263\sqrt{26}

23546\dfrac{2}{3}\sqrt{546}

939\sqrt{3}

Solution:

Let DD be the foot of the perpendicular from the incenter II to BC.BC. The tangent length BD=sAC,BD = s - AC, where s=12(25+26+27)=39,s = \tfrac12(25 + 26 + 27) = 39, so BD=3926=13.BD = 39 - 26 = 13.

By Heron's formula the area is 39141312,\sqrt{39 \cdot 14 \cdot 13 \cdot 12}, and the inradius satisfies r2=(sa)(sb)(sc)s=14131239=56.r^2 = \dfrac{(s-a)(s-b)(s-c)}{s} = \dfrac{14 \cdot 13 \cdot 12}{39} = 56.

In right triangle BDI,BDI, BI2=r2+BD2=56+169=225,BI^2 = r^2 + BD^2 = 56 + 169 = 225, so BI=15.BI = 15.

Thus, the correct answer is A.

Problem 18 in Other Years